91Ó°ÊÓ

An example of a bounded sequence without a limit is an oscillating sequence, where the terms alternate between two values. Consider the sequence \((a_n)_{n \in \mathbb{N}}\) defined by: \[ a_n = (-1)^n \]

To show that the sequence \((a_n)_{n \in \mathbb{N}}\) is bounded, we need to find a lower bound and an upper bound for its terms. The terms of the sequence are alternating between -1 and 1, so we can say that: \[ -1 \leq a_n \leq 1, \quad \forall n \in \mathbb{N} \] Therefore, the sequence \((a_n)_{n \in \mathbb{N}}\) is bounded by -1 and 1.

To prove that the sequence \((a_n)_{n \in \mathbb{N}}\) has no limit, we need to show that there does not exist a limit L for which the terms of the sequence get arbitrarily close to as the index approaches infinity. Suppose, for the sake of contradiction, that the sequence has a limit L. Then, for any \(\varepsilon > 0\), there exists a natural number N such that: \[ |a_n - L| < \varepsilon, \quad \forall n > N \] Now, consider the absolute difference between consecutive terms of the sequence: \[ |a_{n+1} - a_n| = |(-1)^{n+1} - (-1)^n| \] For even values of n (when \(n=2k\) for some \(k\in \mathbb{N}\)), we have: \[ |a_{n+1} - a_n| = |-1 - 1| = 2 \] Similarly, for odd values of n (when \(n=2k+1\) for some \(k\in \mathbb{N}\)), we have: \[ |a_{n+1} - a_n| = |1 - (-1)| = 2 \] From the assumption that the sequence has a limit L, we could find an N such that for all \(n>N\): \[ |a_n - L| < 1 \] Applying this to both \(a_n\) and \(a_{n+1}\), we have: \[ |a_n - L| + |a_{n+1} - L| < 1 + 1 \] \[ |a_{n+1} - L| + |a_n - L| < 2 \] But we know that: \[ |a_{n+1} - a_n| \geq |a_{n+1} - L| + |a_n - L| \] By the triangle inequality, this implies that: \[ 2 \geq |a_{n+1} - a_n| \geq 2 \] This is a contradiction, as we have shown earlier that the absolute difference between consecutive terms of the sequence is always equal to 2. Hence, the sequence \((a_n)_{n \in \mathbb{N}}\) does not have a limit.